Following data are obtained for the reaction
`N_(2)O_(5)to2 NO_(2)+(1)/(2)O_(2)`

(a) Show that it follows first order reaction
(b) Calculate the half-life
(Given `log2=0.3010,log4=0.6021`)

Answer

`t_(1//2)=300s`

Explanation

`[N_(2)O_(5)]_(0)=1.6xx10^(-2)M,k=(2.303)/(t)log""([N_(2)O_(5)]_(0))/([N_(2)O_(5)])`
At `t=300" s",k=(2.303)/(300)log""(1.6xx10^(-2))/(0.8xx10^(-2))=(2.303)/(300)log2=2.31xx10^(-3)`
At `t=600" s",k=(2.303)/(600)xxlog""(1.6xx10^(-2))/(0.4xx10^(-2))=(2.303)/(600)log4=2.31xx10^(-3)`
As k comes out to be constant, it is a reaction of 1st order.
`t_(1//2)=(0.693)/(k)=(0.693)/(2.31xx10^(-3))=300s`
or `1.6xx10^(-2)" M to "0.8xx10^(-2)` M means half of the substance has reacted in 300 s. Henece, `t_(1//2)=300 s.`